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Answer 1 · 2017-03-18T18:28:44

The answer is #=1/2ln(x^2+1)+1/(2x^2+2)+C#

We perform this integration by substitution

Let #u=x^2+1#

#du=2xdx#

#xdx=(du)/2#

Therefore,

#int(x^3dx)/(x^2+1)^2=int(x^2*xdx)/(x^+1)^2#

#=1/2int((u-1)du)/u^2#

#=1/2int(1/u-1/u^2)du#

#=1/2int(du)/u-1/2int(du)/u^2#

#=1/2lnu+1/2*1/u#

#=1/2ln(x^2+1)+1/(2x^2+2)+C#